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3 years ago

Can someone please help me

Physics

1 answer:

mario62 [17]3 years ago

8 0

Answer:

GE = ME - $\frac{1}{2} m\,v^2$ , which agrees with option C in your list.

Explanation:

The definition of Mechanical Energy (ME) of a system is the addition of the gravitational potential energy (GE) plus the kinetic energy (KE) of the system:

ME = GE + KE

Given that the KE is: $\frac{1}{2} m\,v^2$ ,

solving for GE in the formula above gives:

GE = ME - KE = ME - $\frac{1}{2} m\,v^2$ , which agrees with option C

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A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. what is the average power delivered by the engine? (1 hp = 746 w)

AVprozaik [17]

The solution for this is:
Work done = force * distance = m*a*d and power = energy/time
The vo=0 and vf = 25 m/s and t=7 sec. This gives...
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N.
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again).

5 0

3 years ago

Read 2 more answers

A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?

pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, $Y=20\times 10^{10}\ N/m^2$

The young modulus of a wire is given by :

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

$Y=\dfrac{F.L}{A\Delta L}$

$\Delta L=\dfrac{F.L}{A.Y}$

$\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}$

$\Delta L=0.034\ m$

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0

4 years ago

he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug

S_A_V [24]

Answer:

207.4 N

Explanation:

The torque $\tau$ on a body is

$\tau = r* F$ where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

$\tau = rF\sin \theta$

Where $\theta$ is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation $\tau = rF\sin \theta$

$\tau = LF\sin \theta$

Where L is the length of the wrench.

Making F the subject

$F = \frac{\tau }{{L\sin \theta }}$

Force required to pull the wrench is given as,

$F = \frac{\tau }{{L\sin \theta }}$

Substitute $47{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ , 25 cm for L, and 115o for $\theta$

$\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}$

6 0

3 years ago

A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

$d_1 = v* t$

$d_1 = 20 * 0.8$

$d_1 = 16 m$

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2*(-7)*d_2$

$d_2 = 28.6 m$

so the total distance covered by it

$d = d_1 + d_2$

$d = 16 + 28.6 = 44.6 m$

so it will cover a total distance of 44.6 m

3 0

3 years ago

In an electronic transition an atom can not emit what?

iren [92.7K]

█ Question █

In an electronic transition, an atom cannot emit what?

█ Answer █

When an electronic transition is occurring, an atom cannot emit ultra-violet light.

Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia

7 0

3 years ago

Can someone please help me​

Can someone please help me